Exercise 3.1

Question

ewcommand\pptp[1]{(x_{#1}, y_{#1})}

ewcommand\pptt[1]{x_{#1} 	imes y_{#1}}

ewcommand\peq[1]{y_{#1} - x_{#1}^2}

Define two points $\pptp{0}$ and $\pptp{1}$ of the plane to be equivalent if $\peq{0} = \peq{1}$.
Check that this is an equivalence relation and describe the equivalence classes.

Dan Whitman · Accepted Answer

ewcommand\pptp[1]{(x_{#1}, y_{#1})}

ewcommand\pptt[1]{x_{#1} 	imes y_{#1}}

ewcommand\peq[1]{y_{#1} - x_{#1}^2}

First we show that this relation, which we shall denote with $\sim$, is an equivalence relation.
\begin{proof}
  In what follows, suppose that $\pptp{0}$, $\pptp{1}$, and $\pptp{2}$ are all points in the plane.
  
  (Reflexivity) Of course we have $\peq{0} = \peq{0}$, and hence $\pptp{0} \sim \pptp{0}$.

(Symmetry) Suppose that $\pptp{0} \sim \pptp{1}$.
  Then we have $\peq{0} = \peq{1}$ so that of course $\peq{1} = \peq{1}$ since numerical equality is symmetric, and so $\pptp{1} \sim \pptp{0}$ as well.

(Transitivity) Suppose that $\pptp{0} \sim \pptp{1}$ and $\pptp{1} \sim \pptp{2}$.
  Then $\peq{0} = \peq{1}$ and $\peq{1} = \peq{2}$ so that of course $\peq{0} = \peq{2}$ since numerical equality is transitive.
  Therefore $\pptp{0} \sim \pptp{2}$, which shows transitivity.

This suffices to show that $\sim$ is an equivalence relation as we set out to show.
\end{proof}

Each equivalence class formed by this relation is the parabola $y = x^2$ shifted up or down on the $y$-axis.
This is easy to see since two points $\pptp{0}$ and $\pptp{1}$ are in the same class if $\peq{0}$ and $\peq{1}$ have the same value, say $c$.
Then $\peq{0} = c$ so that $y_0 = x_0^2 + c$, which is clearly such a parabola, and similarly $y_1 = x_1^2 + c$.

Exercise 3.1

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Exercise 3.1

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